📚 Year 11 CIE Statistics: Case Study Practice | Year 11 CIE 统计：案例分析实战演练
Real-world data analysis sharpens statistical intuition. This article walks through a complete case study – a survey of 50 students’ study hours and mock exam scores – demonstrating how every CIE IGCSE Statistics topic, from data collection to hypothesis testing and regression, applies in practice.
真实世界的数据分析能磨砺统计直觉。本文通过一个完整案例分析——对50名学生每日学习时间与模拟考试成绩的调查——展示从数据收集到假设检验与回归的每一个 CIE IGCSE 统计知识点如何在实际中运用。
1. Data Collection and Preliminary Sorting | 数据收集与初步整理
A sample of 50 Year 11 students was randomly selected. Each student reported their average daily study time (hours) and their latest mock exam score (%). A clean dataset is the foundation of all analysis; check for missing values and outliers before proceeding.
随机抽取了50名 Year 11 学生作为样本。每名学生报告了平均每日学习时间（小时）以及最近的模拟考试成绩（百分比）。干净的数据集是一切分析的基础，在继续之前要检查缺失值和异常值。
The study-time data ranged from 1.2 h to 9.0 h. One student reported 12.5 h, but after verification it was a recording error and was corrected to 5.2 h. Exam scores were between 38% and 98%.
学习时间数据范围在 1.2 小时到 9.0 小时之间。一名学生报告了 12.5 小时，但经核实是记录错误，已修正为 5.2 小时。考试成绩介于 38% 到 98% 之间。
2. Descriptive Statistics: Measures of Central Tendency | 描述统计：集中趋势量数
For study hours, the mean is 5.24 h, the median is 5.1 h, and the mode is 5.0 h. The close agreement suggests a roughly symmetric distribution. For exam scores, the mean is 72.4%, median 73%, mode 70%.
就学习时间而言，均值为 5.24 小时，中位数为 5.1 小时，众数为 5.0 小时。三者接近表明分布大致对称。考试成绩的均值为 72.4%，中位数为 73%，众数为 70%。
The mean and median for both variables are similar, indicating the central location is well captured by the mean. Use the median if extreme scores skew the mean.
两个变量的均值与中位数都相近，说明均值很好地反映了中心位置。若极端值拉偏均值，则使用中位数。
3. Descriptive Statistics: Measures of Dispersion | 描述统计：离散度
Study hours: standard deviation s = 1.82 h, interquartile range (IQR) = 2.6 h (Q1 = 3.8 h, Q3 = 6.4 h). The IQR is less affected by the corrected outlier. Exam scores: s = 12.7%, IQR = 18%.
学习时间：标准差 s = 1.82 小时，四分位距 (IQR) = 2.6 小时 (Q1 = 3.8 h, Q3 = 6.4 h)。IQR 较少受修正的异常值影响。考试成绩：s = 12.7%，IQR = 18%。
Coefficient of variation for study hours is (1.82/5.24)×100% ≈ 34.7%; for scores (12.7/72.4)×100% ≈ 17.5%. Scores are relatively less variable than study hours.
学习时间的变异系数为 (1.82/5.24)×100% ≈ 34.7%；考试成绩的变异系数为 (12.7/72.4)×100% ≈ 17.5%。考试成绩的相对变异程度小于学习时间。
4. Data Visualisation: Histogram and Box Plot | 数据可视化：直方图与箱线图
A histogram of study hours (bin width = 1 h) shows a unimodal, slightly right-skewed shape. The box plot confirms this, with the median closer to Q1, and one mild outlier at 9.0 h just beyond the upper fence (Q3 + 1.5×IQR = 6.4 + 1.5×2.6 = 10.3 h, so 9.0 h is not an outlier).
学习时间的直方图（组距为1小时）呈现单峰、略右偏的形态。箱线图证实了这点，中位数更靠近 Q1，有一个温和异常值 9.0 小时刚好在上边缘以内（上边缘 = 10.3 h，所以 9.0 h 并非异常值）。
For exam scores, the box plot is symmetric with no outliers. The histogram is bell-shaped, hinting at a normal distribution – a crucial observation for later inferential methods.
考试成绩的箱线图对称且无异常值。直方图呈钟形，暗示正态分布——这对后续推断方法至关重要。
5. Basic Probability Applications | 概率基础应用
If we randomly pick one student, what is the probability that they studied more than 7 hours? From the data, 6 out of 50 students exceeded 7 h, so P(>7 h) = 6/50 = 0.12. Similarly, the probability of scoring above 85% is 9/50 = 0.18.
如果随机选取一名学生，其学习时间超过 7 小时的概率是多少？数据中有 6 名学生超过 7 小时，因此 P(>7 h) = 6/50 = 0.12。同理，考试成绩超过 85% 的概率为 9/50 = 0.18。
We can also find joint probabilities: P(>7 h AND score>85%) = 3/50 = 0.06. This small overlap suggests study time alone doesn’t guarantee top scores.
我们还可以计算联合概率：P(>7 h 且 成绩>85%) = 3/50 = 0.06。这个较小的重叠表明，仅靠学习时间并不保证顶尖成绩。
6. Binomial Distribution Modelling | 二项分布建模
Consider a new survey where each student is asked if they use a revision planner. Suppose historically 40% do. If we sample 20 students, the number X who use a planner follows B(20, 0.4). Probability exactly 10 use a planner: P(X=10) = (20C10)(0.4^10)(0.6^10) ≈ 0.117. Using tables or a calculator is acceptable in exams.
设想一项新调查，询问学生是否使用复习计划本。假设历史数据显示 40% 的人使用。如果我们抽取 20 名学生，使用计划本的人数 X 服从 B(20, 0.4)。恰好有 10 人使用的概率：P(X=10) = (20C10)(0.4^10)(0.6^10) ≈ 0.117。考试中可使用表格或计算器。
The mean of this binomial is np = 20×0.4 = 8, variance np(1-p) = 4.8. This helps predict expected outcomes and variability in repeated samples.
该二项分布的均值为 np = 20×0.4 = 8，方差为 np(1-p) = 4.8。这有助于预测重复抽样下的期望结果与变异性。
7. Normal Distribution Application | 正态分布应用
Exam scores are approximately normal with μ=72.4 and σ=12.7. What proportion of scores lie above 85%? Standardise: Z = (85 – 72.4)/12.7 ≈ 0.99. P(Z>0.99) = 1 – 0.8389 = 0.1611. So around 16.1% score above 85%.
考试成绩近似服从正态分布，μ=72.4，σ=12.7。超过 85% 的比例是多少？标准化：Z = (85 – 72.4)/12.7 ≈ 0.99。P(Z>0.99) = 1 – 0.8389 = 0.1611，因此约 16.1% 的学生成绩超过 85%。
The 90th percentile: Z₀.₉₀ ≈ 1.2816, so score = μ + Zσ = 72.4 + 1.2816×12.7 ≈ 88.7%. Only the top 10% achieve above 88.7%.
第 90 百分位数：Z₀.₉₀ ≈ 1.2816，因此分数 = μ + Zσ = 72.4 + 1.2816×12.7 ≈ 88.7%。只有前 10% 的同学达到 88.7% 以上。
8. Sampling Distribution and Standard Error | 抽样分布与标准误
If we repeatedly draw samples of n=50 from a normal population with μ=72.4 and σ=12.7, the sampling distribution of the sample mean X̄ is also normal with mean μ_X̄ = 72.4 and standard error σ/√n = 12.7/√50 ≈ 1.796.
如果我们从 μ=72.4、σ=12.7 的正态总体中反复抽取容量 n=50 的样本，样本均值 X̄ 的抽样分布也是正态分布，均值为 μ_X̄ = 72.4，标准误为 σ/√n = 12.7/√50 ≈ 1.796。
SE = σ/√n = 12.7/√50 ≈ 1.796
Even if the population were not normal, the Central Limit Theorem ensures X̄ is approximately normal for n=50. This underpins confidence intervals and hypothesis tests.
即使总体不服从正态分布，中心极限定理保证当 n=50 时 X̄ 近似服从正态分布。这是置信区间和假设检验的基础。
9. Confidence Interval Construction | 置信区间构建
Construct a 95% confidence interval for the true mean exam score μ. Since σ is known (we use s=12.7 as an estimate, but assume known), the formula is:
构建真实平均考试成绩 μ 的 95% 置信区间。假设 σ 已知（我们用 s=12.7 作为估计值），公式为：
X̄ ± Z₀.₀₂₅ × (σ/√n)
With X̄=72.4, Z₀.₀₂₅=1.96, σ/√n=1.796, the interval is 72.4 ± 1.96×1.796 = (68.88, 75.92). We are 95% confident that the true mean score lies between 68.9% and 75.9%.
代入 X̄=72.4，Z₀.₀₂₅=1.96，σ/√n=1.796，区间为 72.4 ± 1.96×1.796 = (68.88, 75.92)。我们有 95% 的把握认为真实平均分介于 68.9% 与 75.9% 之间。
If σ had been unknown and we used the sample standard deviation, we would use the t-distribution, but CIE IGCSE focuses on the z-interval when σ is known or when n is large.
如果 σ 未知并使用样本标准差，则需采用 t 分布，但 CIE IGCSE 侧重于 σ 已知或 n 较大时的 z 区间。
10. Hypothesis Testing for a Mean | 均值的假设检验
The school claims the mean exam score is at least 75%. Test at the 5% significance level. H₀: μ = 75, H₁: μ < 75 (one-tailed). Test statistic: Z = (72.4 - 75) / 1.796 = -2.6 / 1.796 ≈ -1.447.
学校声称平均考试成绩至少为 75%。在 5% 显著性水平下进行检验。H₀: μ = 75，H₁: μ < 75（单尾）。检验统计量：Z = (72.4 - 75) / 1.796 = -2.6 / 1.796 ≈ -1.447。
Critical value at 5% left-tail: Z_crit = -1.645. Since -1.447 > -1.645, we do not reject H₀. There is insufficient evidence to say the mean is below 75%.
5% 左尾临界值：Z_crit = -1.645。因为 -1.447 > -1.645，我们不拒绝 H₀。没有足够证据表明平均分低于 75%。
The p-value = P(Z < -1.447) ≈ 0.074, which is >0.05, confirming the conclusion.
p 值 = P(Z < -1.447) ≈ 0.074，大于 0.05，证实了结论。
11. Correlation and Regression Analysis | 相关与回归分析
A scatter plot of study hours (x) against exam scores (y) shows a moderate positive linear trend. The Pearson correlation coefficient r is calculated as 0.62, indicating a fairly strong positive linear relationship.
学习时间 (x) 与考试成绩 (y) 的散点图呈现中等正线性趋势。皮尔逊相关系数 r 计算得 0.62，表明较强的正线性关系。
The least-squares regression line is y = 48.5 + 4.57x. For every additional hour of study, the predicted exam score increases by about 4.57 percentage points. The coefficient of determination R² = 0.3844, so about 38.4% of the variation in scores is explained by study time.
最小二乘回归线为 y = 48.5 + 4.57x。每多学习 1 小时，预测考试成绩提升约 4.57 个百分点。决定系数 R² = 0.3844，即考试成绩中约 38.4% 的变异可由学习时间解释。
Always check the residual plot for randomness to validate the linear model. Predict with caution outside the data range (extrapolation).
务必检查残差图的随机性以验证线性模型。在数据范围之外预测（外推）需谨慎。
12. Writing a Statistical Report and Conclusion | 撰写统计报告与结论
A complete case study report should summarise the problem, describe the sampling method, present descriptive statistics and graphs, and interpret inferential results. Always relate findings back to the context – for instance, while study time significantly predicts scores, other factors like sleep and stress also play roles.
一份完整的案例分析报告应当概述问题、描述抽样方法、呈现描述统计量及图表，并解读推断结果。始终将发现联系回实际情境——例如，尽管学习时间能显著预测成绩，睡眠和压力等其他因素也起作用。
Use clear, non-technical language when communicating to non-statisticians. Include confidence intervals and p-values with careful explanations of what they mean (not just ‘significant’).
向非统计人员传达时，使用清晰的非技术语言。在报告中包含置信区间和 p 值，并仔细解释它们的含义（而非仅仅说“显著”）。
This case study illustrates how the CIE IGCSE Statistics toolkit comes together – from raw data to a meaningful conclusion. The same logical flow applies to any investigation, whether in business, biology, or social science.
本案例分析展示了 CIE IGCSE 统计工具箱如何从原始数据出发，最终得出有意义的结论。同样的逻辑流程适用于任何调查研究，无论是在商业、生物学还是社会科学领域。
Published by TutorHao | Statistics Revision Series | aleveler.com
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